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Question
Consider N = n1n2 identical cells, each of emf ε and internal resistance r. Suppose n1 cells are joined in series to form a line and n2 such lines are connected in parallel.
The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n1 and n2 can be continuously varied, find the relation between n1, n2, R and r for which the current in R is maximum.
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Solution
(a)
Given:-
Emf of one cell = E
∴ Total e.m.f. of n1 cells in one row = n1E
Total emf of one row will be equal to the net emf across all the n2 rows because of parallel connection.
Total resistance in one row = n1r
Total resistance of n2 rows in parallel \[= \frac{n_1 r}{n_2}\]
Net resistance of the circuit = R + \[\frac{n_1 r}{n_2}\]
\[\therefore \text{Current, } I = \frac{n_1 E}{R + \frac{n_1 r}{n_2}} = \frac{n_1 n_2 E}{n_2 R + n_1 r}\]
(b) From (a),
\[I = \frac{n_1 n_2 E}{n_2 R + n_1 r}\]
For I to be maximum, (n1r + n2R) should be minimum
\[\Rightarrow \left( \sqrt{n_1 r} - \sqrt{n_2 R} \right)^2 + 2\sqrt{n_1 R n_2 r} = \min\]
It is minimum when
\[\sqrt{n_1 r} = \sqrt{n_2 R}\]
\[ n_1 r = n_2 R\]
∴ I is maximum when n1r = n2R .
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